Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int LVL = 18;
const int N = 1e5 + 100;
vector<int> go[N][LVL];
int p[N][LVL];
int lvl[N];
vector<int> g[N];
vector<int> res;
inline void dfs(const int &v, const int &par) {
  p[v][0] = par;
  lvl[v] = lvl[par] + 1;
  for (int i = 1; i < LVL; ++i) p[v][i] = p[p[v][i - 1]][i - 1];
  for (int i = 1; i < LVL; ++i) {
    go[v][i] = go[v][i - 1];
    for (auto it : go[p[v][i - 1]][i - 1]) go[v][i].emplace_back(it);
    sort(go[v][i].begin(), go[v][i].end());
    go[v][i].resize(min((int)go[v][i].size(), 10));
  }
  for (int to : g[v]) {
    if (to == par) continue;
    dfs(to, v);
  }
}
inline bool bit(const int &mask, const int &pos) { return (mask >> pos) & 1; }
inline int get_p(int v, const int &step) {
  for (int i = 0; i < LVL; ++i)
    if (bit(step, i)) v = p[v][i];
  return v;
}
inline int lca(int a, int b) {
  if (lvl[a] < lvl[b]) swap(a, b);
  a = get_p(a, lvl[a] - lvl[b]);
  if (a == b) return lvl[a];
  for (int i = LVL - 1; i >= 0; --i)
    if (p[a][i] != p[b][i]) a = p[a][i], b = p[b][i];
  a = p[a][0];
  return lvl[a];
}
inline void get_here(int a, const int &step) {
  for (int i = 0; i < LVL; ++i) {
    if (bit(step, i)) {
      for (auto it : go[a][i]) res.emplace_back(it);
      sort(res.begin(), res.end());
      res.resize(min((int)res.size(), 10));
      a = p[a][i];
    }
  }
}
inline void get_ans(int a, int b) {
  int lvl_lca = lca(a, b);
  get_here(a, lvl[a] - lvl_lca + 1);
  get_here(b, lvl[b] - lvl_lca);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    int c;
    cin >> c;
    go[c][0].emplace_back(i);
    sort(go[c][0].begin(), go[c][0].end());
    go[c][0].resize(min((int)go[c][0].size(), 10));
  }
  dfs(1, 0);
  while (q--) {
    int v, u, a;
    cin >> v >> u >> a;
    res.clear();
    get_ans(v, u);
    res.resize(min((int)res.size(), a));
    cout << res.size() << " ";
    for (auto it : res) cout << it << " ";
    cout << "\n";
  }
  return 0;
}